Answer:-
This question is solved by using the simple formula of concentration unit of ppm i.e. parts per million which is also 1 mg/L = 1 ppm.
The answer is given in the image,
What weight of Pb(NO3)2 will have to be dissolved in 1.00 liter of water to prepare...
Calculate the concentration of IO−3IO3− in a 9.88 mM Pb(NO3)29.88 mM Pb(NO3)2 solution saturated with Pb(IO3)2Pb(IO3)2. The ?sp of Pb(IO3)2Pb(IO3)2 is 2.5×10−13. Assume that Pb(IO3)2Pb(IO3)2 is a negligible source of Pb2+Pb2+ compared to Pb(NO3)2Pb(NO3)2. [IO−3]= A different solution contains dissolved NaIO3NaIO3. What is the concentration of NaIO3NaIO3 if adding excess Pb(IO3)2(s)Pb(IO3)2(s) produces a Pb2+Pb2+ concentration of 9.40×10−6 M?
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Calculate the concentration of IO3 in a 1.25 mM Pb(NO3)2 solution saturated with Pb(IO3)2. The Ksp of Pb(IO3)2 is 2.5 x 10-13. Assume that Pb(IO3), is a negligible source of Pb2+ compared to Pb(NO3)2. [103] = 2 x10-10 M A different solution contains dissolved NalO3. What is the concentration of NaIO, if adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 7.00 x 10-6 M? concentration: -1.396 x10-5 M
Calculate the concentration of IO3 in a 1.81 mM Pb(NO3)2 solution saturated with Pb(IO3)2 The Ksp of Pb(IO3)2 is 2.5 x 1013. Assume that Pb(IO3)2 is a negligible source of Pb2 compared to Pb(NO3)2 Number 10;]- М A different solution contains dissolved NalO3. What is the concentration of NalO3 if adding excess Pb(IO3)2(s) produces [Pb21] 5.10 x10 M? Number [Nalo,j= М
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balance the following oxidation reduction reaction in basic
solution NO3- + Pb —-> NO + Pb 2+
Balance basic oxidation-reduction reactions Question Balance the following oxidation-reduction reaction in basic solution. NO; + Pb → NO + Pb2+ Provide your answer below: No; + Pb+ 4,0-Ono + Pb2+ + oH-